Sending one broken taxi each to three different airports. 
Suppose we have $9$ taxis, and three airports, aiport $A$, airport $B$, and airport $C$. We want to send $3$ taxis to airport $A$, $5$ taxis to airport $B$, and $1$ taxi to airport $C$. If exactly three of the taxis are in need of repair, what is the probability
  that every airport receives one of the taxis requiring repairs?

I'm struggling to solve this problem. There is a theorem that I want to use, listed below:

THEOREM: The number of ways of partitioning $n$ distinct objects into $k$ distinct groups containing $n_1, n_2, \ldots , n_k$ objects, respectively, where each object appears in exactly one group and $\sum_{i=1}^k n_i = n$, is $$N = {n\choose n_1 \;n_2 \;\cdots\; n_k} = {n!\over n_1!\;n_2!\;\cdots\;n_k!}.$$

The following is a solution I found here on page 83 for this problem. 
SOLUTION: Using the theorem, there are $${9\choose 3\;5\;1} = {9!\over 3!\;5!\;1!} = 504\text{ ways}$$ to send the taxis to all the airports. Let $W$ denote the event that the $3$ taxis that are in need of repair are sent to each airport. The number of ways we can send each of the $3$ broken taxis to the three airports is given by $${3\choose 1\;1\;1} = 3! = 6. \tag{Why is this important?}$$ Also, the number of ways we can send the remaining $6$ taxis to the three airports is given by $${6\choose 2\;4\;0} = {6!\over2!\;4!} = 15$$ (since each of the three broken taxis are already taking up a spot). So, by the $mn$-rule, the number of points in the sample space for $W$ is given by $$N_W = \underbrace{6\times15}_\text{Why?} = 90.$$ Thus the probability that each of the three airports receives a broken taxi is $$P(\text{broken taxi to each } A,B,C) = {N_W\over N} = {90\over504}.$$

Can someone please explain the two Whys? I thought that by holding each of the three broken taxis constant that we would only need to consider the remaining $6$, and I originally thought the probability would be $15\over504$. However, this probability seems unusually low. Thus, I went and found the solution written above. However, I was not able to comprehend the logic behind the solution. 
Here's what I'm thinking is wrong with my logic process and how we get the correct logic: Holding each of the broken taxis to a particular airport is where the logic is incorrect. We also want to be able to permute the broken taxis among the three airports. This is why we need the ${3\choose1\;1\;1}$. We can then permute the remaining $6$ taxis among the three airports, giving us the ${6\choose 2\;4\;0}$. Is this correct thinking?

An earlier problem asked us to find the probability that exactly one broken taxi is sent to airport $C$. I did the following approach:
If we're sending $1$ broken taxi to airport $C$, then we can permute the remaining $8$ like so $${8\choose 3 \; 5\; 0} = 56,$$ which this answer was correct according to the back of the book. However, I applied the same principle to the problem at-hand and only received $15$ ways. Is the "correct" way to use the theorem and the $mn$-rule to say: There is $${1\choose 0\;0\;1} = 1$$ way to send $1$ broken taxi to airport $C$, and $56$ ways to distinctly send the other $8$ to airports $B$ and $C$. So, by the $mn$-rule, there are $1\times56$ sample points such that a broken taxi gets sent to airport $C$. This makes sense to me, but I might just be making up a solution that yields the correct answer, but is not logically correct.
 A: Without using fancy formulas you can get the answer by simple enumeration.
There are 3 broken taxis that need to be distributed amongst 3 airports.  We have been given how many taxis that need to go to each airport.
By simply distributing the broken taxis like the below:
3 to the A, 0 to B and 0 to C:  this can be done in ${3\choose3}{5\choose0}{1\choose0} = 1$
2 to A, 1 to B and 0 to C : ${3\choose3}{5\choose0}{1\choose0} = 15$
2 to A, 0 to B and 1 to C : ${3\choose2}{5\choose0}{1\choose1} = 3$
1 to A, 2 to B and 0 to C : ${3\choose1}{5\choose2}{1\choose0} = 30$
1 to A, 1 to B and 1 to C : ${3\choose1}{5\choose1}{1\choose1} = 15$
0 to A, 3 to B and 0 to C : ${3\choose0}{5\choose3}{1\choose0} = 10$
0 to A, 2 to B and 1 to C : ${3\choose0}{5\choose2}{1\choose1} = 10$
Total would be 84 (${9\choose3}$.
What you want is 1 to each airport and that is 15.
Thus the required probability is $\frac{15}{84} = \frac{90}{504}$ from your method.
Thanks
A: The first Why can be interpreted as: there are 6 different ways to send the 3 broken taxis so that each of them end up at different airport. For example, if your broken taxis are 2, 3, 7, then taxi 2 can go to airport B, taxi 3 to airport C, and taxi 7 to airport A. It can easily be seen that there are 6 different arrangements for this scenario, even just by permutation (3 airports for taxi 2 to end up, then 2 airports for taxi 3, then 1 airport for taxi 7).
The second Why can be interpreted exactly as you mentioned: holding each of the three broken taxis constant (meaning we know exactly which airport they ended up), how many taxi-to-airport arrangements there are for the remaining 6 working taxis.
Lastly, since there are 6 different ways that the 3 broken taxis can end up (as seen from the first Why), each with 15 different ways that the remaining 3 working taxis can end up, the total number of arrangements will be 6 * 15 = 90. In other words, the answer of 15/504 is wrong because it does not take into account the 6 different arrangements that the bad taxis can end up.
